हिंदी

Abcd is a Quadrilateral in Which Ad = Bc. E, F, G and H Are the Mid-points of Ab, Bd, Cd and Ac Respectively. Prove that Efgh is a Rhombus - Mathematics

Advertisements
Advertisements

प्रश्न

ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.

योग
Advertisements

उत्तर

Given that AD = BC                                                    …..(1)

From the figure,
For triangle ADC and triangle ABD

2GH = AD and 2EF = AD, therefore 2GH = 2EF = AD …..(2)

For triangle BCD and triangle ABC

2GF = BC and 2EH=BC, therefore 2GF= 2EH = BC      …..(3)

From (1), (2) ,(3) we get,
2GH = 2EF = 2GF = 2EH
GH = EF = GF = EH
Therefore EFGH is a rhombus.
Hence proved.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Mid-point and Its Converse [ Including Intercept Theorem] - Exercise 12 (A) [पृष्ठ १५०]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 12 Mid-point and Its Converse [ Including Intercept Theorem]
Exercise 12 (A) | Q 8 | पृष्ठ १५०

संबंधित प्रश्न

ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.


In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.


In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.


The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

PR = `[1]/[2]` ( AB + CD)


In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R.
Prove that : (i) AP = 2AR
                   (ii) BC = 4QR


In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR


ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.


In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔHEB ≅ ΔHFC


P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.


E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).

[Hint: Join BE and produce it to meet CD produced at G.]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×